Odds of Winning Mega Millions

A lottery draw is a selection without replacement where order does not matter, so the number of possible tickets is a binomial coefficient:

$\binom{n}{k} = \frac{n!}{k!\,(n-k)!}$

Mega Millions draws 5 white balls from 70 and 1 Mega Ball from a separate pool of 24 (the Mega Ball pool was 25 before the April 2025 format change — the method is identical, only the multiplier changes).

$\binom{70}{5} = \frac{70 \times 69 \times 68 \times 67 \times 66}{5 \times 4 \times 3 \times 2 \times 1} = 12{,}103{,}014$

Because the Mega Ball comes from an independent pool, multiply:

$N = 12{,}103{,}014 \times 24 = 290{,}472{,}336$

Each ticket is equally likely, so $P(\text{jackpot}) = 1/N$. Assumptions: every combination is equally likely, the two pools are independent, and your numbers are fixed before the draw.

To count tickets that match exactly $k$ of your 5 white balls, choose which of your numbers hit and which of the 65 non-drawn numbers fill the rest:

$\text{ways} = \binom{5}{k}\binom{65}{5-k}$

Multiply by $1$ if the Mega Ball must match, or by $23$ if it must not. Then

$P = \frac{\binom{5}{k}\binom{65}{5-k} \times (1 \text{ or } 23)}{290{,}472{,}336}$

Odds vs probability. Casinos and lotteries quote "1 in $N$" — that is $1/N$, a probability. True odds against would be $N-1$ to $1$; the difference is negligible at these magnitudes but matters in a stats exam.

Independence across draws. Tickets in the same draw with different numbers are mutually exclusive, so buying $t$ distinct tickets gives $P = t/N$ exactly. Playing the same numbers across $t$ different draws gives $1 - (1 - 1/N)^t$, which is very slightly less.

• Using permutations. $70 \times 69 \times 68 \times 67 \times 66$ counts ordered draws and overcounts by $5! = 120$. Divide by $120$.
• Adding the Mega Ball pool instead of multiplying. Two independent stages multiply: $\binom{70}{5} \times 24$.
• Counting "at least $k$" as "exactly $k$". $\binom{5}{3}\binom{65}{2}$ excludes tickets that match 4 or 5; add those tiers separately if the question says "at least".
• Forgetting the 23 wrong Mega Balls. Matching 5 whites without the Mega Ball has 23 ways, not 1.
• Believing hot or cold numbers. Draws are independent; past frequencies carry no information.
• Ignoring jackpot sharing. Expected value must divide the jackpot by the expected number of co-winners, which grows with ticket sales.